comparison src/fftw-3.3.3/kernel/buffered.c @ 10:37bf6b4a2645

Add FFTW3

+/*
+ * Copyright (c) 2003, 2007-11 Matteo Frigo
+ * Copyright (c) 2003, 2007-11 Massachusetts Institute of Technology
+ *
+ * This program is free software; you can redistribute it and/or modify
+ * it under the terms of the GNU General Public License as published by
+ * the Free Software Foundation; either version 2 of the License, or
+ * (at your option) any later version.
+ *
+ * This program is distributed in the hope that it will be useful,
+ * but WITHOUT ANY WARRANTY; without even the implied warranty of
+ * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
+ * GNU General Public License for more details.
+ *
+ * You should have received a copy of the GNU General Public License
+ * along with this program; if not, write to the Free Software
+ * Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA  02110-1301  USA
+ *
+ */
+
+/* routines shared by the various buffered solvers */
+
+#include "ifftw.h"
+
+#define DEFAULT_MAXNBUF ((INT)256)
+
+/* approx. 512KB of buffers for complex data */
+#define MAXBUFSZ (256 * 1024 / (INT)(sizeof(R)))
+
+INT X(nbuf)(INT n, INT vl, INT maxnbuf)
+{
+     INT i, nbuf, lb; 
+
+     if (!maxnbuf) 
+	  maxnbuf = DEFAULT_MAXNBUF;
+
+     nbuf = X(imin)(maxnbuf,
+		    X(imin)(vl, X(imax)((INT)1, MAXBUFSZ / n)));
+
+     /*
+      * Look for a buffer number (not too small) that divides the
+      * vector length, in order that we only need one child plan:
+      */
+     lb = X(imax)(1, nbuf / 4);
+     for (i = nbuf; i >= lb; --i)
+          if (vl % i == 0)
+               return i;
+
+     /* whatever... */
+     return nbuf;
+}
+
+#define SKEW 6 /* need to be even for SIMD */
+#define SKEWMOD 8 
+
+INT X(bufdist)(INT n, INT vl)
+{
+     if (vl == 1)
+	  return n;
+     else 
+	  /* return smallest X such that X >= N and X == SKEW (mod SKEWMOD) */
+	  return n + X(modulo)(SKEW - n, SKEWMOD);
+}
+
+int X(toobig)(INT n)
+{
+     return n > MAXBUFSZ;
+}
+
+/* TRUE if there exists i < which such that maxnbuf[i] and
+   maxnbuf[which] yield the same value, in which case we canonicalize
+   on the minimum value */
+int X(nbuf_redundant)(INT n, INT vl, int which, 
+		      const INT *maxnbuf, int nmaxnbuf)
+{
+     int i;
+     (void)nmaxnbuf; /* UNUSED */
+     for (i = 0; i < which; ++i)
+	  if (X(nbuf)(n, vl, maxnbuf[i]) == X(nbuf)(n, vl, maxnbuf[which]))
+	       return 1;
+     return 0;
+}